Chebyshev Inequality bounds for the probability of 90 out of 100 coin tosses being tails The Chebyshev Inequality states: $$P(|X - \mu| \ge c) \le \frac{\sigma^2}{c^2}$$
My understanding of this is that this means that the inequality gives us the upper bounds on the probability that the random variable will differ from the mean ($\mu$) by at least $c$.
I am trying to use that to calculate the bounds of the probability that, if I flip a fair coin 100 times, I'll get tails at least 90 times. In particular, I am interested in whether or not the Chebyshev Inequality would identify this an an anomaly. (In my previous question, which this question is related to, I used the Markov Inequality for that purpose; the answer and comments showed me that I was correct in stating that the Markov Inequality would not identify it as an anomaly).
Since I'm calculating the probability of a specific number trials "succeeding," I thought that $\sigma^2$ should be the variance of the Binomial Distribution. I calculated the variance of the Binomial Distribution as follows: $$np(1-p)=(100)(0.5)(1 - 0.5)=25$$
Clearly, the expected number of "tails" is 50 (i.e. half heads and half tails), so $\mu = 50$. This means that $c = 90 - 50 = 40$, so I calculated that $$P(|X - \mu| \ge c) \le \frac{25}{40^2} \approx 0.016$$ So the probability that there are at least 90 or at most 10 tails is at most 1.6%. The probability that there were at least 90 tails is half of that (approximately 0.8%).
This is quite unsurprising in one sense because I was expecting this to be an anomalous event; however, what is more surprising is how much this differs from Markov Inequality. (I do understand from the previous Q&A that the Markov Inequality is a pretty weak bound, but I wasn't expecting the Chebyshev Inequality to be that much stronger).
Can someone help me see whether I did this correctly and if my understanding is correct, or if I'm missing anything?
 A: This is correct. Another way to phrase the same bound: since $\sigma^2 = 25$, the standard deviation is $\sigma =5$, so $90$ is $8$ standard deviations away from the mean. The probability of being at least $8$ standard deviations away from the mean is at most $\frac1{8^2} = \frac1{64}$.
This is not actually all that strong when compared to the true probability of about $3.06 \times 10^{-17}$. (That's the two-sided probability, which you should compare to $\frac1{64}$.) It's much better than nothing, though!
In general, Chebyshev's inequality could be compared to a normal approximation, because both of them are bounds on how many standard deviations we are from the mean. Also, for the coin flip example, the normal approximation is very accurate.
For a random variable with mean $\mu$ and standard deviation $\sigma$:

*

*Chebyshev's inequality tells us nothing about $\Pr[|X-\mu| > \sigma]$. For a normal distribution, this probability is about $31.7\%$.

*Chebyshev's inequality says $\Pr[|X - \mu| > 2\sigma]$ is at most $\frac14 = 25\%$. For a normal distribution, this probability is about $4.55\%$.

*Chebyshev's inequality says $\Pr[|X-\mu| > 3\sigma]$ is at most $\frac19 \approx  11.1\%$. For a normal distribution, this probability is about $0.27\%$.

*In the long run, Chebyshev's inequality says that $\Pr[|X-\mu| > k \sigma]$ is at most $\frac1{k^2}$. For a normal distribution, this probability decays very approximately as $e^{-k^2/2}$: exponentially faster.

(However, for large values of $k$, the normal approximation will not be particularly accurate, even for a random variable that's usually close to being normal. Use Chernoff-type inequalities to get the exponentially decaying tail bounds in such cases.)
